Question Number 88503 by jagoll last updated on 11/Apr/20
$$\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$
Commented by john santu last updated on 11/Apr/20
![cos ((7π)/8)=cos (π−(π/8))= −cos (π/8) cos ((5π)/8) =−cos ((3π)/8) ⇒(1+cos (π/8))(1−cos (π/8))= 1−cos^2 (π/8)=sin^2 ((π/8)) ⇒(1+cos ((3π)/8))(1−cos ((3π)/8))=1−cos^2 (((3π)/8))=sin^2 (((3π)/8)) ∴ [ sin ((3π)/8).sin (π/8)]^2 = (1/4)[ cos ((4π)/8)−cos (π/4)]^2 = (1/4)×(1/2)=(1/8)](Q88508.png)
$$\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{8}}\right)=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:=−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)=\:\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}=\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\therefore\:\left[\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}.\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\right]^{\mathrm{2}} =\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{8}}−\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\right]^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\: \\ $$
Commented by peter frank last updated on 11/Apr/20
$${thank}\:{you} \\ $$
Commented by peter frank last updated on 15/Apr/20
$${help}\:{Qn}\:\mathrm{88937} \\ $$